I will assume that this is a question where we use the central limit theorem toOriginally Posted bykingkaisai2

justify treating the distribution of the mean score as normal with mean equal

to the mean for a single throw of the die and variance equal to the variance

of the result of a single throw divided by the sample size .

a) Now a mean of less than 3.3 means 230 or less for the sum, so to correct

for the continuity of the normal approximation we use a z-score for a mean

of 230.5/70, which is the z-score for for a mean of 3.2928 so:

Looking this up in a normal table or using a normal calculator we have for

a normal distributed RV we have:

b) There are at least two equivalent ways to do this, one another is to

observe that the event: "total score exceeds 260.5" is the same as "the mean

exceeds 260.5/70". I will use the latter, then the z-score for a mean of

260.5/70 (again the extra 0.5s are continuity corrections) is:

Looking this up as before we find:

so:

.

RonL

Notes:

i. A mean of less than 3.3 means the sum of the 70 throws is less than

3.3x70=231. So as the normal distribution is continuous and the actual

distribution of the sum and mean is discrete we use in the normal model a

sum less than 230.5 to represent an actual sum of 230 or less.

ii. Similarly in part b), a total of more than 260 is modelled as more than 260.5

in the normal approximation.

iii. Given the nature of the way problems are set, there is a significant chance

that the expected answers to the two parts of this problem are the same.

So please check the working to see if there is some slight modification which

would make this so.